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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 16

  • question_answer
    If \[y=a\ln x+b{{x}^{2}}+x\]has its extreme values at x= -1,2 then a + b=

    A) 2         

    B) \[\frac{3}{2}\]    

    C) 1                     

    D) \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

    [b]:\[y'=\frac{a}{x}+2bx+1,y'(-1)=0,y'(2)=0\] \[\Rightarrow \]\[-a-2b+1=0,\frac{a}{2}+4b+1=0\] \[\Rightarrow \]\[a=2,b=-\frac{1}{2}\Rightarrow a+b=\frac{3}{2}\].


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